Clare Snyder writes:

On Sun, 4 Apr 2021 21:19:14 -0500, Robert Nichols
wrote:

On 4/4/21 4:52 PM, Clare Snyder wrote:
On Sun, 4 Apr 2021 09:36:53 -0500, Robert Nichols
wrote:

On 4/3/21 11:47 PM, Clare Snyder wrote:
On Sat, 3 Apr 2021 19:38:26 -0700 (PDT), Linda Iverson
wrote:

On Friday, August 2, 2019 at 6:18:04 AM UTC-7, Goncalves wrote:
replying to mjacobsen925, Goncalves wrote:
Really mjacobsen925, a forum is where one asks questions not where one gets
censored.
--
for full context, visit https://www.polytechforum.com/metalw...pe-639655-.htm

What if I just want to use black steel pipe as a curtain rod to hang heavy velvet drapes spanning a window greater than 12 feet long without the need for a center support? Ill anchor ends using flanges (nipples and elbows) screwed into solid wooden header above window. Seems like it should be more than strong enough for such a girly thing.
2 or 3 inch might work but not 3/4 or 1 inch unless you want them to
sag

The pipe, even 2 inch, will sag just from its own weight.
Sched 40 will sag - sched 80 MIGHT work in 2 inch but not with HEAVY
drapery

A 12 foot length of 2 inch Sched 80 steel pipe would weigh over 60 lbs.
That's quite some curtain rod!!
But it will support itself and some load over a 12 ft span. (about 1
1/2 times?? as much as sched 40.

Just did as bit of investigating and 1 1/4 inch sched 80 should be
adequate for a pretty heavy curtain - and weigh about 40 lb? - about
10 lb per meter) will sipport about 30 lb evenly didtributed with
about an inch of deflection.
Sched 40 will only handle around 20 lb with almost 1 1/2 inches of
deflection with a wight of ropughly 30 lb.

That's Euler-Bernoulli beam - very clear and exact. You can have the
load it can take and the deflection at that load to great accuracy.
If you set-up the load on the finished article, you'd find the
deflection matched to within a millimetre or something like that.
The end supports cannot be anything like rigid enough against rotation
to benefit the load capacity and stiffness against deflection.
You've got a "simple supported beam" ("double-supported beam").
A fair and reasonable conservative assumption would be to put the
entire weight of the curtain in the middle of the "curtain rail" when
doing the Euler-Bernoulli calculation for what tube to specify.
So that truly is the "simply supported centrally-loaded beam" case.

I do a lot of these calculations.
eg.
http://www.weldsmith.co.uk/tech/stru...lat_calcs.html


Rich Smith